We know that if we have, for example, exponential function in the numerator and square function in the denominator, and both approach infinity, then the limit is infinity since exponential function approaches infinity faster.
Let's consider
$\lim\limits_{x \to \infty} {e^x} × \frac{1}{x}$
I know this aprroaches infinity since we can write it as ${e^x}/{x}$ but as I can't think of a better example this has to do it.
My question is can we conclude that limit is infinity without any transformations, in general case, if multiplicand aproaches infinity faster then multiplier approaches $0$?
Best Answer
Take the definition of the limit. You want to show that, for any $A > 0$, there exists $x_0$ such that $x > x_0 \Rightarrow f(x) > A$.
Now, if I understand correctly, you sak whether you can split the problem when $f(x) = g(x) \times h(x)$ and study separately each, with $g(x)$ going to infinity.
If you can bound $h(x)$ from below, $h(x)>\varepsilon>0$, knowing $g$ goes to infinity you can find an $x_0$ from which $g(x) > \frac{A}{\varepsilon}$ and so $f(x) > A$.
But if $h(x)$ tends to zero, then you cannot make this lower bound on $h(x)$ and thus need to compare $g$ and $h$ pointwise, so make the product in some way.